Question

Natural gas consists of 60.0 mole% ethane and 40.0 mole% butane. The gas is stored in a 300 ft^3 tank at a pressure of 1500 psig, and a temperature of 220 F and sold at a cost of $0.15/lb.

a) Determine whether the ideal gas equation of state is a good approximation for the given conditions.

b) Estimate the cost to fill the tank using the compressibility-factor equation of state.

Answer 1

Temperature of the gas mixture = 220 deg.F, from F= 1.8C+32, C= 104.4 deg.c, P= 1500Psig= 1500+14.7 =1514.7 Psia = 1514.7/14.7 =103.0408 atm.

Ideal gas law is obeyed at low pressure and high enough temperature, where the attractive forces between the gas moecules can be neglcted. Here that is not the case. So ideal gas law cannot be used.

b Critical priperties

Ethane : 32.2 deg.c = 32.2+273= 363.1K, Pressure = 709 Psia

butane : 152 deg.c= 152+273= 425K, pressure = 550 Psia

Tr= T/TC and Pr=P/PC

for ethane, Tr= (104.4+273)/363.1 =1.04, Pr = 1514.7/709 = 2.13

Butane, Tr= (104.4+273)/425 = 0.888, Pr= 1514.7/550 = 2.754

Compressibility factors ( from charts)

ethane=0.35 and butane = 0.21

So compressibility of mixture = 0.35*0.6+0.21*0.4 = 0.294

hence from gas law equation,PV= ZRnT

n= numbef of moles, P= Pressrue = 103.04 atm, V= Volume = 300ft3= 300*28.32 L=8496L, T= 104.4+273= 377.4 K

R= 0.082 L.atm/mole.K, Z =0.294

n= PV/ZRT = 103.04*8496/(0.294*377.4*0.0821) =96101 moles

gmoes Moles of each gas : ethane = 0.6* 96101=57661, butane= 38440

Mass = moles* molar mass , molar masses : ethane = 30 , butane = 58

Masses : Ethane = 57661*30 =1729830 gm = 1730 kg butane = 38440*58=1922000= 1922 Kg

total mass= 1730+1922= 3652 kg

0.4535 kg = 1lb

3652 kg = 3652/0.4535=8053 lb

1 lb costs 0.15$

8053 lb costs 8053*0.15=1208$